The Deligne Dam on the Cayley River is built so that the wall facing the water i
ID: 2866092 • Letter: T
Question
The Deligne Dam on the Cayley River is built so that the wall facing the water is shaped like the region above the curve y=0.7x^2 and below the line y=220. (Here, distances are measured in meters.) The water level is 36 meters below the top of the dam. Find the force (in Newtons) exerted on the dam by water pressure. (Water has a density of 1000kg/m^3, and the acceleration of gravity is 9.8m/sec^2 .) i need to know how to do this so best explanation will be rewarded points. thank you. btw i have posted this question twce and both answers were wrong. the answer is not 115920000. I am adding more points to the question in hope of the correct answer. thank you
Explanation / Answer
Height = 220 - 36 = 184
y = 0.7x^2
y/0.7 = x^2
10y/7 = x^2
x = +/- sqrt(10y/7)
So the length of a thin horizontal strip is : 2*sqrt(10y/7)
The force on that strip is the pressure due to a weight of water above that height, y, in a 1m by 1m column, times that area.
So, Area = 2sqrt(10y/7)*dy
The force on that strip is the pressure due to a weight of water above that height, y, in a 1m by 1m column, times that area.
The weight of that water is mgh= (9.8)(1000)(184 - y) so your integral is :
(integral from 0 to 184) 2sqrt(10y/7)*9.8*1000*(184 - y)*dy
Taking constants out :
19600*sqrt(10/7) * (int from 0 to 184) (184 - y)*sqrt(y)*dy
19600*sqrt(10/7) * (int from 0 to 184) (184y^(1/2) - y^(3/2))*dy
19600*sqrt(10/7) * (int from 0 to 184) (184*2/3*y^(3/2) - (2/5)y^(5/2)
Plug in limits :
19600*sqrt(10/7) * ((184*2/3*184^(3/2) - (2/5)184^(5/2) - 0
19600*sqrt(10/7)*122465.3674179675083325
2868932571.495816088882268188
So, it is 2868932571.50 Newton ---> ANSWER
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